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Inversion transform with center (or pole) $C$ and power $k^2$ is defined by:

$$\tag{1}J_{C,k}:M \leftrightarrow M' \ \ \ \ \ \iff \ \ \ \ \ \ \ \vec{CM'}=\frac{k^2}{||\vec{CM}||^2} \ \vec{CM} $$

It is an "involutive" transform: $M'$ is the image of $M$ iff $M$ is the image of $M'$. This explains the double arrow.

This transform, presented for the first time in a systematic way by Plücker, (1834) is detailed in many books/sites.

See Appendix 1 for a comment about inversion properties displayed on Fig. 1.

enter image description here

Fig. 1. : Inversion with center $C$ and power 1. Shapes in black are exchanged by inversion with shapes in red. The blue circle, called "circle of inversion", is the locus of invariant points.

But there are ill-known features that I would like to gather.

Moreover, I would like to enlarge this "quest" to interesting (not fully standard) questions/applications of inversion.

For the initialization of this "collection", I propose 3 themes and a compendium of interesting facts:

1) "Champagne glass inversion" : Fig. 2 shows how inversion with center $C$ and radius of inversion $k$ (we find back the large and the small fish, inverted images one of the other...) can be realized by a 3 steps $P\Gamma P^{-1}$ operation with respect to a paraboloid $\Pi$, where:

  • $P$ is the vertical projection from $\Pi$ onto the horizontal plane, and

  • $\Gamma$ is the conical projection from $\Pi$ to $\Pi$ with center $C_0(a,b,a^2+b^2-k^2)$ where $(a,b)$ are the coordinates of $C$.

This non-classical way to describe inversion deserves an explanation that we have placed in Appendix 2.

enter image description here

Fig. 2. Inversion by a conical projection $\Pi \to \Pi$.

2) "Bubble inversion" This is a cousin representation of the first one with a sphere instead of a parabola (see Fig. 3). It necessitates to use 2 steps with the two stereographic projections $S_N$ and $S_S$ from the North and South pole resp. with respect to the equatorial plane. Take a look at Fig. 3.

Consider a certain figure, say the circle in red. It is transformed by $S_N$ into the circle on the unit sphere materialized by black little diamonds ; this circle is transformed in turn by $S_S$ into the blue circle on the equatorial plane... which is the inversive image of the initial red circle. Briefly said :

$$\tag{$\star$}I=S_S \circ S_N$$

This shouldn't come as a surprize because stereographic projections are themselves 3D inversions.

See (composition of stereographic projections is inversion through the ball - a geometric way) for a proof.

enter image description here

Fig. 3. (planar) inversion realized by combining two stereographic projections (formula ($\star$)).

3) "How inversion transform is connected with linear algebra" (in fact connected with item 1) ; this issue looks paradoxical because inversion is definitely not a linear transform. In fact, there exists a $4 \times 4$ linear representation of the co-called conformal geometry including another category of non-linear transforms, translations. Reference: the very good book "Riemannian Geometry" by S. Gallot, D. Hullin, J. Lafontaine, 2nd edition 1993. Universitext, Springer, pages 175-176. Here is how this correspondence is done (Explanations will be found in this book or in the very interesting development by @MvG cited in 4a):

$$\text{If} \ J \ \text{is the basic inversion (center 0, power 1):}$$ $$[J]:=\begin{pmatrix}0&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{pmatrix}.$$ $$\text{If} \ H_r \ \text{is the homothety with ratio} \ r \ \text{and center} \ 0: $$ $$[H_r]:=\begin{pmatrix}1&0&0&0\\0&r&0&0\\0&0&r&0\\0&0&0&r^2\end{pmatrix}.$$

$$\text{If} \ T_V \ \text{is the translation by vector} \ V=\binom{a}{b} :$$ $$[T_V]:=\begin{pmatrix}1&0&0&0\\2a&1&0&0\\2b&0&1&0\\a^2+b^2&a&b&1\end{pmatrix}.$$

All these transformations preserve the following quadratic form:

$$q(x_0;x_1,x_2,x_3):=x_1^2+x_2^2-x_0x_3$$

where $x_0$ should be considered as a homogeneous coordinate (i.e., is there to take into account the projective dimension). The signature of $q$ being $(+++-)$, we are dealing with elements of the group classically denoted $O(3,1)$.

This correspondence with 2D transforms can be extended in a straightforward way to higher dimensions (for $n$D transforms, the correspondence is with $(n+2) \times (n+2)$ matrices).

4) Complex representation with an application :

As recalled by @José Carlos Santos, inversion with the unit circle $S^1=\{z\in\mathbb{C}\,|\,|z|=1$} as invariant circle is the map

$$I : \begin{cases}\ \begin{array}{ccc}\mathbb{C}\setminus\{0\}&\longrightarrow&\mathbb{C}\setminus\{0\}\\z&\mapsto&\frac{1}{\overline z}\end{array}\end{cases}$$

Let us prove the following result :

Any (2D) rotation can be obtained as a certain combination of translations and inversions.

Let us consider the following identity (as given in https://mathoverflow.net/q/19965) :

$$e^{i \theta} + \frac{1}{-e^{-i \theta} + \frac{1}{e^{i \theta} + \frac{1}{z}}} = - e^{2 i \theta} z. \tag{%}$$

If we conjugate both sides of (%), we get :

$$e^{-i \theta} + \frac{1}{\overline{-e^{-i \theta} + \frac{1}{\overline{e^{-i \theta} + \frac{1}{\overline{z}}}}}} = - e^{-2 i \theta}\overline{z}. \tag{%%}$$

(The LHS of (%%) can be written in a symbolic way as : $T_{e^{-i \theta}}\circ I \circ T_{-e^{-i \theta}} \circ I \circ T_{e^{-i \theta}} \circ I$).

If we take $\theta=0$ in (%%), one gets :

$$1+ \frac{1}{\overline{-1 + \frac{1}{\overline{1 + \frac{1}{\overline{z}}}}}} = - \overline{z}. \tag{%%%}$$

(Please note the conjugation bar). (%%%) means that transformation $z\mapsto-\overline{z}$ can be obtained by combining only inversions and translation(s).

In a similar manner (%%) means that transformation $z\mapsto -e^{-2i\theta}\overline{z}$, can obtained using as well inversions and translations.

Combining both transformations, we can thus transform $z$ into $e^{-2i\theta}z$ for any theta, using translations and inversions only.

5) A compendium of interesting documents/issues

My (enlarged) question is : can you give examples (of your own) of interesting, possibly non-usual, uses of inversion ?

Appendix 1: Recall of properties of inversion as depicted by Fig. 1:

  • The image of a straight line is in general a circle that passing through the pole $C$, and, exceptionally, when the straight line passes through the origin, its image is the line itself (caution: in the latter case, the straight line is "globally" invariant, but its points, in general, aren't invariant).

  • The image $\Gamma'$ of a circle $\Gamma$ not passing through the origin is a circle of the same type (caution: the center of $\Gamma'$ is not the image of the center of $\Gamma$ ; nevertheless these centers are aligned with the pole of inversion). The image of a circle passing through the origin is a straight line, as said before.

  • The image of the black fish made of line segments is the red fish made of circular arcs swimming in its blue round fishbowl.

Appendix 2: Explanation of inversion depicted in Fig. 2.

Let us begin by another definition of inversion.

$$\tag{3}J_{C,k}:M \to M' \ \ \ \ \iff \ \ \ \ \ \exists \vec{U} \ \text{(unit norm vector) s.t.} \ \begin{cases}\vec{CM}=\lambda \vec{U}\\ \vec{CM'}=\lambda' \vec{U}\end{cases} \ \text{and} \ \lambda\lambda'=k^2$$

Let $M(x,y)$ and $M'(x',y')$ exchanged by inversion $J_{C,k}$.

Let $M_0(x,y,x^2+y^2)$ and $M_0'(x',y',x'^2+y'^2)$ be their "lifted version" on paraboloid $\Pi$.

Let $N(X,Y,Z) $ be any point of line $C_0M_0$. We can define an abscissa $\mu$ on this line for point $N$ in the following way:

$$\tag{$\star$}\vec{C_0N}=\mu \vec{C_0M_0} \ \ \ \ \iff \ \ \ \ \begin{cases}X=a+\mu(x-a)\\ Y=b+\mu(y-b)\\ Z=c+\mu(x^2+y^2-c)\end{cases}$$

with $$c:=a^2+b^2-k^2.$$

Please note that for $\mu=1$, $N$ is in $M_0$. Let $\lambda'$ be the value of $\mu$ associated with $M'_0$.

Thus $N \in \Pi$ in two cases: when $\mu=1$ ($N=M_0$) and when $\mu=\lambda'$ ($N=M'_0$). As

$$\tag{3}N \in \Pi \ \iff \ Z=X^2+Y^2,$$

plugging in (3) the expressions of $X,Y,Z$ in $(\star)$, and replacing $c$ by its defining expression, we obtain the following quadratic equation with unknown $\mu$:

$$\mu^2[(x-a)^2+(y-b)^2] \ + \ \mu[\cdots] \ + \ k^2=0$$

Using the classical formula for the roots' product, we have

$$\lambda' \times 1 =\dfrac{k^2}{(x-a)^2+(y-b)^2}=\dfrac{k^2}{\|\vec{CM}\|^2}$$

We find back here the definition of inversion given in (2) (recall that $\vec{CM}$ hasn't been normalized).

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An application of the inversion to mechanical linkages (converting circular to linear motion): Peaucellier–Lipkin linkage.

Sylvester writes that when he showed a model to (Lord) Kelvin, he “nursed it as if it had been his own child, and when a motion was made to relieve him of it, replied ‘No! I have not had nearly enough of it—it is the most beautiful thing I have ever seen in my life.’”

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  • $\begingroup$ Thank you for reminding me this beautiful, and ill known indeed, way of doing inversion (and the funny quote). $\endgroup$ – Jean Marie Jan 29 '18 at 21:13
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Chapter 8 of Evan Chen's book 'Euclidean Geometry in Mathematical Olympiads' has a pretty neat application of inversion to the Shoemaker's Knife Problem. For more information see - http://www.math.ubc.ca/~cass/courses/m308-03b/projects-03b/hunter/hunter.html

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Perhaps this is not what you are interested in, but here it goes: the inversion with respect to the circle $S^1=\{z\in\mathbb{C}\,|\,|z|=1$} is simply the map$$\begin{array}{ccc}\mathbb{C}\setminus\{0\}&\longrightarrow&\mathbb{C}\setminus\{0\}\\z&\mapsto&\overline z^{\,-1}.\end{array}$$

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  • $\begingroup$ It's in the scope of my question : aspects of inversion that don't come immediately to mind when you see the definition... $\endgroup$ – Jean Marie Jan 29 '18 at 11:55
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Several properties of the circles in a Pappus chain can be derived using circle inversion, as this picture borrowed from wikipedia shows:

enter image description here

Along the same idea, a construction based on inversion has been proposed for drawing the figure with TikZ in an answer to The Pappus Chain on tex.stackexchange.

The Pappus chain is also related to Steiner chains, which are in turn related to Soddy's hexlet and the Dupin cyclides mentioned by the OP.

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Any four distinct points A, B, C, D can be inverted into the vertices of a parallelogram $A'B'C'D'$ (including, as one possibility, a degenerate parallelogram in which the four vertices lie on one line, but still $A'B' = D'C'$ and $A'D' = B'C'$). Hint: Consider separately the three cases (i) $AC // BD$; (ii) $AB // CD$ or $AD // BC$; (iii) A, B, C, D not concyclic.

Note: the // notation refers to point-pair separation. One way of characterizing is that AC // BD when every circle through A and C intersects every circle through B and D.

Finally I actually asked a question Is this usage of the term 'Jacobian' related to the other uses e.g. matrix or elliptic function? related to this (where this is repeated) that I never had anyone answer!

The diagrams below give my rendition of the book (Geometry Revisited) solution for the (iii) case.

Exercise 5.7.6

5.7.6 supplementary

5.7.6 Lemma

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  • $\begingroup$ Interesting property I wasn't aware of ! $\endgroup$ – Jean Marie Feb 28 at 10:59

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